document.write( "Question 1121413: The ATM at a local convenience store allows customers to make withdrawals of $10, $20, $50, or $100. Let X denote a random variable that indicates the amount withdrawn by a customer. The probability distribution of X is
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document.write( "P(10) = 0.2
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document.write( "P(20) = 0.5
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document.write( "P(50) = 0.2
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document.write( "P(100) = 0.1
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document.write( "(a) Draw the probability distribution of X.
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document.write( "(b) What is the probability that a customer withdraws more than $20?
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document.write( "(c) What is the expected amount of money withdrawn by a customer?
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document.write( "(d) The expected value is not a possible value of the amount withdrawn. Interpret the expected value for a manager.
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document.write( "(e) Find the variance and standard deviation of X. \n" );
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Algebra.Com's Answer #737310 by Boreal(15235)![]() ![]() You can put this solution on YOUR website! Prob >$20 is 0.3 \n" ); document.write( "E(X)=0.2*10+0.5*20+0.2*50+0.1*100=$32 \n" ); document.write( "The expected value may be considered to be a weighted average, how much will be withdrawn on average. In other words, the expected value*the number of people withdrawing will be about what the total amount of money needed is. \n" ); document.write( "variance is $22^2*0.2+$12^2*0.5+$18^2*0.2+$88^2*0.1=$96.80+$72+$64.80+$774.40=$1008 \n" ); document.write( "sd is sqrt (V)=$31.75 \n" ); document.write( " |